In classA the average is the quiecent (bias) current.
With more input it shift to classB and the current get clamped on one side to ground (can't go lower) but the other side goes up and so does the average.
Now if you clamp the up-going side to two times the quiecent current then you take away the same amount on both sides and the average is the quiescent current again.
Mona
With more input it shift to classB and the current get clamped on one side to ground (can't go lower) but the other side goes up and so does the average.
Now if you clamp the up-going side to two times the quiecent current then you take away the same amount on both sides and the average is the quiescent current again.
Mona
Yes it works for class AB.
I read through the article. I have two questions:
1) The idle current is controlled by Rk and the voltage drop across the two diodes. Si diode has Temp Co of -2mV/deg C. A change of 20 deg C ( which is very possible as the tube heat up ), that's -80mV decrease across the resistor Rk. If you assume 0.56V per diode drop, That is about 7.1% drop in current during warmup. So the circuit introduce 7% error with temperature just from the tube heating up if assuming 20 deg rise in the environment.
2) According to my understanding, the circuit using C1=C2 and R1=R2 to assume both get pumped up the same amount during a signal appeared at the input. But C2 is charged by R2 through a diode, C1 is charged and discharged through R1 directly. The two path is not the same, C1 is charge and discharge bi-directionally, C2 only get charge through the diode, there is no discharge path when the diode turns off during the negative swing. That is: low going swing at the cathode will discharge C1 through R1. But the low going signal will turn off the diode and will not discharge C2. I would think voltage across C2 will get higher than C1 during any input signal. Am I missing the point here?
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